L'Hôpital's Rule for Indeterminate Limits

When a limit takes an indeterminate form such as $0 / 0$ or $\pm \infty / \pm \infty$ , L'Hôpital's Rule says to differentiate the numerator and denominator separately and retry. The same idea extends to products, differences, and exponential forms by first converting them to a ratio.

Indeterminate form	Technique
$0 / 0$ or $\pm \infty / \pm \infty$	Apply L'Hôpital's Rule directly
$0 \cdot (\pm \infty)$	Rewrite as $f / (1 / g)$ or $g / (1 / f)$
$\infty - \infty$	Use common denominator, rationalize, or factor
$0^{0}$ , $1^{\pm \infty}$ , $(\pm \infty)^{0}$	Take $\ln$ , reduce to $0 \cdot (\pm \infty)$

L'Hôpital's Rule for $0 / 0$

Assume $f (a) = g (a) = 0$ . For $x \neq a$ :

\frac{f (x)}{g (x)} = \frac{f (x) - f (a)}{g (x) - g (a)} = \frac{\frac{f (x) - f (a)}{x - a}}{\frac{g (x) - g (a)}{x - a}} .

If f'(a) and g'(a) both exist and g'(a)\neq0, taking the limit as $x \to a$ gives:

\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f'(a)}{g'(a)}.\tag{i}

Two panels showing graphs of f and g near x=a, demonstrating that zooming in makes them look linear and the ratio approaches f'(a)/g'(a) — Zooming in near $x = a$ , the graphs of $f$ and $g$ look nearly linear, so their ratio approaches f'(a)/g'(a).

L'Hôpital's Rule for $0 / 0$ . Assume

lim_{x \to s} f (x) = 0 and lim_{x \to s} g (x) = 0,

and that \lim_{x\to s}f'(x)/g'(x) exists (or is $+ \infty$ or $- \infty$ ). Then:

\lim_{x\to s}\frac{f(x)}{g(x)}=\lim_{x\to s}\frac{f'(x)}{g'(x)}.

Here $s$ denotes $a$ , $a^{+}$ , $a^{-}$ , $- \infty$ , or $+ \infty$ .

Proof

For

x \to a^{+}

, introduce extended functions: F(x)=\begin{cases}f(x)&x\neq a\\0&x=a\end{cases},\quad G(x)=\begin{cases}g(x)&x\neq a\\0&x=a\end{cases}. For

a < x

, apply Cauchy's Mean Value Formula to

[a, x]

: there exists

c \in (a, x)

such that f(x)g'(c)=g(x)f'(c), giving f(x)/g(x)=f'(c)/g'(c). As

x \to a^{+}

c \to a^{+}

, so \lim_{x\to a^+}f(x)/g(x)=\lim_{c\to a^+}f'(c)/g'(c). For

x \to + \infty

, the substitution

x = 1 / u

reduces the problem to

u \to 0^{+}

Important notes:

f'(x)/g'(x) is the ratio of the derivatives, not the derivative of the ratio.
Apply L'Hôpital's Rule only when the limit takes the indeterminate form $0 / 0$ . If it is not $0 / 0$ , substituting directly is correct.
Write $\overset{H}{=}$ above the equals sign to indicate an application of L'Hôpital's Rule.

To apply L'Hôpital's Rule:

Verify that the limit takes the form $0 / 0$ (or $\pm \infty / \pm \infty$ ).
Differentiate $f (x)$ and $g (x)$ separately (not the ratio).
Find \lim_{x\to s} f'(x)/g'(x). If it is a number, $+ \infty$ , or $- \infty$ , that is the answer. Otherwise, you cannot conclude the original limit does not exist, but you cannot use L'Hôpital's Rule further in this form.
If the result is still $0 / 0$ or $\pm \infty / \pm \infty$ , repeat.

Stop differentiating as soon as the numerator or denominator is no longer zero (or infinite) at $s$ .

Examples: Form $0 / 0$

Find $lim_{x \to 0} \frac{\sin x}{1 - e^{x}}$ .

Solution

f (0) = \sin 0 = 0

and

g (0) = 1 - e^{0} = 0

, so we apply L'Hôpital's Rule:

lim_{x \to 0} \frac{\sin x}{1 - e^{x}} \overset{H}{=} \frac{\cos 0}{- e^{0}} = \frac{1}{- 1} = - 1.

Find $lim_{x \to 1} \frac{1 - x^{3}}{1 - x}$ .

Solution

lim_{x \to 1} \frac{1 - x^{3}}{1 - x} \overset{H}{=} lim_{x \to 1} \frac{- 3 x^{2}}{- 1} = 3.

Find $lim_{x \to 1} \frac{x^{3} - 3 x + 2}{2 x^{3} - x^{2} - 4 x + 3}$ .

Solution

lim_{x \to 1} \frac{x^{3} - 3 x + 2}{2 x^{3} - x^{2} - 4 x + 3} \overset{H}{=} lim_{x \to 1} \frac{3 x^{2} - 3}{6 x^{2} - 2 x - 4} .

x = 1

: still

0 / 0

. Apply L'Hôpital's Rule again:

\overset{H}{=} lim_{x \to 1} \frac{6 x}{12 x - 2} = \frac{6}{10} = \frac{3}{5} .

**Warning:** After the second application, the limit

6 x / (12 x - 2)

is no longer indeterminate at

x = 1

. Do not apply L'Hôpital's Rule a third time.

Find $lim_{t \to 2} \frac{\sqrt{t + 7} - 3}{t - 2}$ .

Solution

lim_{t \to 2} \frac{\sqrt{t + 7} - 3}{t - 2} \overset{H}{=} lim_{t \to 2} \frac{\frac{1}{2 \sqrt{t + 7}}}{1} = \frac{1}{2 \sqrt{9}} = \frac{1}{6} .

Find $lim_{x \to 0} \frac{1 - \cos x}{x^{2}}$ .

Solution

lim_{x \to 0} \frac{1 - \cos x}{x^{2}} \overset{H}{=} lim_{x \to 0} \frac{\sin x}{2 x} \overset{H}{=} lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2} .

Find $lim_{x \to 0} \frac{x - \sin x}{x - \tan x}$ .

Solution

lim_{x \to 0} \frac{x - \sin x}{x - \tan x} \overset{H}{=} lim_{x \to 0} \frac{1 - \cos x}{1 - \sec^{2} x} .

Simplify algebraically:

\frac{1 - \cos x}{1 - \sec^{2} x} = \frac{1 - \cos x}{\frac{\cos^{2} x - 1}{\cos^{2} x}} = \frac{\cos^{2} x (1 - \cos x)}{(\cos x - 1) (\cos x + 1)} = \frac{- \cos^{2} x}{\cos x + 1} .

lim_{x \to 0} \frac{- \cos^{2} x}{\cos x + 1} = - \frac{1}{2} .

Find $lim_{x \to 1} \frac{\ln x}{x - 1}$ .

Solution

lim_{x \to 1} \frac{\ln x}{x - 1} \overset{H}{=} lim_{x \to 1} \frac{1 / x}{1} = 1.

Find $lim_{x \to 0} \frac{x - \sin x}{\sinh x}$ .

Solution

lim_{x \to 0} \frac{x - \sin x}{\sinh x} \overset{H}{=} lim_{x \to 0} \frac{1 - \cos x}{\cosh x} = \frac{1 - 1}{1} = 0.

Graph of y = (x - sin x) / sinh x showing the function approaching 0 at the origin

Find $lim_{x \to 0} \frac{\arcsin 4 x - 2 \arcsin 2 x}{x^{3}}$ .

Solution

The limit is

0 / 0

. Using

\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}}

lim_{x \to 0} \frac{\arcsin 4 x - 2 \arcsin 2 x}{x^{3}} \overset{H}{=} lim_{x \to 0} \frac{\frac{4}{\sqrt{1 - 16 x^{2}}} - \frac{4}{\sqrt{1 - 4 x^{2}}}}{3 x^{2}} .

Still

0 / 0

. Writing the numerator as

4 (1 - 16 x^{2})^{- 1 / 2} - 4 (1 - 4 x^{2})^{- 1 / 2}

and differentiating:

\overset{H}{=} lim_{x \to 0} \frac{64 (1 - 16 x^{2})^{- 3 / 2} - 16 (1 - 4 x^{2})^{- 3 / 2}}{6} = \frac{64 - 16}{6} = 8.

Find $lim_{x \to \infty} \frac{e^{1 / x^{2}} - 1}{2 \arctan x^{2} - π}$ .

Solution

x \to \infty

e^{1 / x^{2}} \to e^{0} = 1

and

\arctan x^{2} \to π / 2

, so the limit is

0 / 0

. Applying L'Hôpital's Rule:

lim_{x \to \infty} \frac{e^{1 / x^{2}} - 1}{2 \arctan x^{2} - π} \overset{H}{=} lim_{x \to \infty} \frac{- 2 x^{- 3} e^{1 / x^{2}}}{4 x / (1 + x^{4})} = lim_{x \to \infty} \frac{- (1 + x^{4})}{2 x^{4}} e^{1 / x^{2}} = - \frac{1}{2} .

Find $lim_{x \to + \infty} \frac{x^{- 5 / 3}}{\sin (1 / x)}$ .

Solution

x \to + \infty

, both

x^{- 5 / 3} \to 0

and

\sin (1 / x) \to 0

, giving

0 / 0

lim_{x \to + \infty} \frac{x^{- 5 / 3}}{\sin (1 / x)} \overset{H}{=} lim_{x \to + \infty} \frac{- \frac{5}{3} x^{- 8 / 3}}{- \frac{1}{x^{2}} \cos (1 / x)} = lim_{x \to + \infty} \frac{5 x^{- 2 / 3}}{3 \cos (1 / x)} = \frac{0}{3 (1)} = 0.

Evaluate $lim_{x \to - \infty} \frac{\sin (1 / x)}{\ln \frac{x + 1}{x + 2}}$ .

Solution

x \to - \infty

\sin (1 / x) \to 0

and

\ln \frac{x + 1}{x + 2} \to \ln 1 = 0

, giving

0 / 0

lim_{x \to - \infty} \frac{\sin (1 / x)}{\ln \frac{x + 1}{x + 2}} \overset{H}{=} lim_{x \to - \infty} \frac{- \frac{1}{x^{2}} \cos \frac{1}{x}}{\frac{1}{(x + 2) (x + 1)}} = lim_{x \to - \infty} (- \frac{(x + 1) (x + 2)}{x^{2}} \cos \frac{1}{x}) = - (1) (1) = - 1.

Limitations of L'Hôpital's Rule

If $f, g \to 0$ as $x \to s$ , but \lim_{x\to s}f'(x)/g'(x) does not exist (and is not $\pm \infty$ ), we cannot conclude that $lim_{x \to s} f (x) / g (x)$ fails to exist. For example, with $f (x) = x^{2} \sin (1 / x)$ and $g (x) = x$ : f'(x)/g'(x)=2x\sin(1/x)-\cos(1/x), which does not have a limit as $x \to 0$ (because $\cos (1 / x)$ oscillates). Yet:

lim_{x \to 0} \frac{x^{2} \sin (1 / x)}{x} = lim_{x \to 0} x \sin \frac{1}{x} = 0.

Also, sometimes L'Hôpital's Rule leads to circular or increasingly complex expressions. Examples:

$lim_{x \to 0^{+}} \frac{e^{- 1 / x}}{x}$ : repeated application gets more complicated; substitute $t = 1 / x$ instead to get $lim_{t \to + \infty} \frac{t}{e^{t}} \overset{H}{=} 0$ .
$lim_{x \to + \infty} \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$ : L'Hôpital's Rule cycles back to the original expression. Factor out $e^{x}$ instead: $lim_{x \to + \infty} \frac{1 - e^{- 2 x}}{1 + e^{- 2 x}} = 1$ .

L'Hôpital's Rule for $\pm \infty / \pm \infty$

L'Hôpital's Rule for $\pm \infty / \pm \infty$ . Assume $lim_{x \to s} f (x) = \pm \infty$ and $lim_{x \to s} g (x) = \pm \infty$ . If \lim_{x\to s}f'(x)/g'(x)=L (or $\pm \infty$ ), then:

lim_{x \to s} \frac{f (x)}{g (x)} = L .

Find $lim_{x \to π / 2^{+}} \frac{\tan 5 x}{\tan x}$ .

Solution

Both

\tan 5 x

and

\tan x \to - \infty

x \to (π / 2)^{+}

, giving

\infty / \infty

lim_{x \to π / 2^{+}} \frac{\tan 5 x}{\tan x} \overset{H}{=} lim_{x \to π / 2^{+}} \frac{5 \sec^{2} (5 x)}{\sec^{2} x} = lim_{x \to π / 2^{+}} \frac{5 \cos^{2} x}{\cos^{2} (5 x)} \overset{0 / 0}{\overset{H}{=}} lim_{x \to π / 2^{+}} \frac{\sin 2 x}{\sin 10 x} \overset{H}{=} lim_{x \to π / 2^{+}} \frac{2 \cos 2 x}{10 \cos 10 x} = \frac{2 \cos 0}{10 \cos 0} = \frac{1}{5} .

Evaluate $lim_{x \to + \infty} \frac{\ln x}{x^{α}}$ where $α > 0$ .

Solution

This is

\infty / \infty

lim_{x \to + \infty} \frac{\ln x}{x^{α}} \overset{H}{=} lim_{x \to + \infty} \frac{1 / x}{α x^{α - 1}} = lim_{x \to + \infty} \frac{1}{α x^{α}} = 0.

This shows that

x^{α}

grows much faster than

\ln x

for any

α > 0

Evaluate $lim_{x \to 0^{+}} \frac{\ln x}{\csc x}$ .

Solution

\ln x \to - \infty

and

\csc x \to + \infty

x \to 0^{+}

, giving

- \infty / \infty

lim_{x \to 0^{+}} \frac{\ln x}{\csc x} \overset{H}{=} lim_{x \to 0^{+}} \frac{1 / x}{- \csc x \cot x} = lim_{x \to 0^{+}} \frac{- \sin^{2} x}{x \cos x} \overset{H}{=} lim_{x \to 0^{+}} \frac{- 2 \sin x \cos x}{\cos x - x \sin x} = 0.

Indeterminate Form $0 \cdot (\pm \infty)$

If $f (x) \to 0$ and $g (x) \to \pm \infty$ as $x \to s$ , rewrite the product as a fraction:

f (x) g (x) = \frac{f (x)}{1 / g (x)} or \frac{g (x)}{1 / f (x)},

producing the form $0 / 0$ or $\pm \infty / \pm \infty$ , then apply L'Hôpital's Rule.

Evaluate $lim_{x \to π / 2} (\sec 3 x \cos 5 x)$ .

Solution

x \to π / 2

\cos 5 x \to 0

and

\sec 3 x \to \pm \infty

, giving

0 \cdot (\pm \infty)

. Write:

\sec 3 x \cos 5 x = \frac{\cos 5 x}{\cos 3 x} = [\frac{0}{0}] .

lim_{x \to π / 2} \frac{\cos 5 x}{\cos 3 x} \overset{H}{=} lim_{x \to π / 2} \frac{- 5 \sin 5 x}{- 3 \sin 3 x} = \frac{5 \sin (5 π / 2)}{3 \sin (3 π / 2)} = \frac{5 (1)}{3 (- 1)} = - \frac{5}{3} .

Graph of y = sec(3x)cos(5x) near x = pi/2, showing the function approaching -5/3

Evaluate $lim_{x \to π / 2} (π - 2 x) \tan x$ .

Solution

(π - 2 x) \to 0

and

\tan x \to \pm \infty

x \to π / 2

. Write as

0 / 0

lim_{x \to π / 2} (π - 2 x) \tan x = lim_{x \to π / 2} \frac{π - 2 x}{\cot x} \overset{H}{=} lim_{x \to π / 2} \frac{- 2}{- (1 + \cot^{2} x)} = \frac{- 2}{- (1 + 0)} = 2.

Graph of y = (pi - 2x) tan x near x = pi/2, approaching 2

Evaluate $lim_{x \to 0^{+}} x \ln \sin x$ .

Solution

x \to 0

and

\ln \sin x \to - \infty

, giving

0 \cdot (- \infty)

lim_{x \to 0^{+}} x \ln \sin x = lim_{x \to 0^{+}} \frac{\ln \sin x}{1 / x} = [\frac{- \infty}{+ \infty}] \overset{H}{=} lim_{x \to 0^{+}} \frac{(\cos x) / \sin x}{- 1 / x^{2}} = lim_{x \to 0^{+}} (- x^{2} \cot x) .

= lim_{x \to 0^{+}} (\cos x) (\frac{x}{\sin x}) (- x) = (1) (1) (0) = 0.

Graph of y = x ln(sin x) near x = 0, approaching 0

Evaluate $lim_{x \to - \infty} x e^{x}$ .

Solution

e^{x} \to 0

x \to - \infty

, giving

0 \cdot (- \infty)

lim_{x \to - \infty} x e^{x} = lim_{x \to - \infty} \frac{x}{e^{- x}} \overset{H}{=} lim_{x \to - \infty} \frac{1}{- e^{- x}} = 0.

Graph of y = x e^x, approaching 0 as x approaches negative infinity

Indeterminate Form $\infty - \infty$

To evaluate a limit of the form $\infty - \infty$ , convert to a fraction using a common denominator, rationalization, or factoring, then apply L'Hôpital's Rule.

Evaluate $lim_{x \to π / 2} (\sec x - \tan x)$ .

Solution

Using a common denominator:

\sec x - \tan x = \frac{1 - \sin x}{\cos x} = [\frac{0}{0}] .

lim_{x \to π / 2} \frac{1 - \sin x}{\cos x} \overset{H}{=} lim_{x \to π / 2} \frac{- \cos x}{- \sin x} = \frac{0}{1} = 0.

Evaluate $lim_{x \to 0} (\frac{1}{\sin x} - \frac{1}{x})$ .

Solution

lim_{x \to 0} \frac{x - \sin x}{x \sin x} \overset{H}{=} lim_{x \to 0} \frac{1 - \cos x}{\sin x + x \cos x} \overset{H}{=} lim_{x \to 0} \frac{\sin x}{2 \cos x - x \sin x} = \frac{0}{2} = 0.

Evaluate $lim_{x \to 1} (\frac{1}{\ln x} - \frac{1}{x - 1})$ .

Solution

lim_{x \to 1} \frac{x - 1 - \ln x}{(x - 1) \ln x} \overset{H}{=} lim_{x \to 1} \frac{1 - 1 / x}{\ln x + 1 - 1 / x} \overset{H}{=} lim_{x \to 1} \frac{1 / x^{2}}{1 / x + 1 / x^{2}} = \frac{1}{1 + 1} = \frac{1}{2} .

Indeterminate Forms $0^{0}$ , $1^{\pm \infty}$ , $(\pm \infty)^{0}$

Limits of the form $lim_{x \to s} f (x)^{g (x)}$ are indeterminate when:

$f \to 0$ and $g \to 0$ (form $0^{0}$ ),
$f \to 1$ and $g \to \pm \infty$ (form $1^{\pm \infty}$ ),
$f \to \pm \infty$ and $g \to 0$ (form $(\pm \infty)^{0}$ ).

Strategy: Let $y = f (x)^{g (x)}$ , take $\ln y = g (x) \ln f (x)$ (form $0 \cdot \pm \infty$ ), find $lim \ln y = L$ , then $lim y = e^{L}$ .

Note: $0^{+ \infty} = 0$ and $0^{- \infty} = + \infty$ are not indeterminate.

Evaluate $lim_{x \to 0^{+}} x^{x}$ .

Solution

Let

y = x^{x}

, so

\ln y = x \ln x

. As

x \to 0^{+}

, this is

0 \cdot (- \infty)

lim_{x \to 0^{+}} x \ln x = lim_{x \to 0^{+}} \frac{\ln x}{1 / x} \overset{H}{=} lim_{x \to 0^{+}} \frac{1 / x}{- 1 / x^{2}} = lim_{x \to 0^{+}} (- x) = 0.

Therefore

\ln y \to 0

, so

y = e^{\ln y} \to e^{0} = 1

Evaluate $lim_{x \to + \infty} x^{1 / x}$ .

Solution

Form

(\infty)^{0}

. Let

y = x^{1 / x}

, so

\ln y = (\ln x) / x

lim_{x \to + \infty} \frac{\ln x}{x} \overset{H}{=} lim_{x \to + \infty} \frac{1 / x}{1} = 0.

\ln y \to 0

and

y \to e^{0} = 1

Evaluate $lim_{x \to 0} (1 + \sin x)^{1 / x}$ .

Solution

Form

1^{\infty}

. Let

y = (1 + \sin x)^{1 / x}

, so

\ln y = \frac{\ln (1 + \sin x)}{x}

lim_{x \to 0} \frac{\ln (1 + \sin x)}{x} \overset{H}{=} lim_{x \to 0} \frac{\cos x / (1 + \sin x)}{1} = 1.

\ln y \to 1

and

y \to e^{1} = e

Graph of y = (1 + sin x)^(1/x), approaching e at x = 0

Evaluate $lim_{x \to 0^{+}} (\cot x)^{\sin x}$ .

Solution

Form

\infty^{0}

. Let

y = (\cot x)^{\sin x}

, so

\ln y = \sin x \ln \cot x

lim_{x \to 0^{+}} \frac{\ln \cot x}{1 / \sin x} \overset{H}{=} lim_{x \to 0^{+}} \frac{(1 + \cot^{2} x) (- 1 / \cot x)}{\cos x / \sin^{2} x} = lim_{x \to 0^{+}} \frac{\sin x}{\cos x} = 0.

\ln y \to 0

and

y \to e^{0} = 1

Frequently Asked Questions

When should I NOT use L'Hôpital's Rule?

Do not apply L'Hôpital's Rule if the limit is not in an indeterminate form. For example,

lim_{x \to 0} (1 + 2 x) / (1 - x) = 1

by simple substitution. Applying L'Hôpital's Rule there (incorrectly) gives

2 / (- 1) = - 2

, which is wrong.

Can L'Hôpital's Rule be applied repeatedly?

Yes, as long as each application produces another indeterminate form. Stop as soon as the limit can be evaluated by substitution. Never apply L'Hôpital's Rule when the result is not indeterminate.

What if L'Hôpital's Rule does not resolve the limit?

L'Hôpital's Rule can fail to resolve the limit in two ways: (1) the ratio of derivatives cycles back to the original form (as in

lim_{x \to + \infty} (e^{x} - e^{- x}) / (e^{x} + e^{- x})

), or (2) the ratio of derivatives has no limit (as in the

x^{2} \sin (1 / x) / x

example). In these cases, use algebraic manipulation, substitution, or other techniques to evaluate the limit directly.

L'Hôpital's Rule for $0 / 0$

Examples: Form $0 / 0$

Limitations of L'Hôpital's Rule

L'Hôpital's Rule for $\pm \infty / \pm \infty$

Indeterminate Form $0 \cdot (\pm \infty)$

Indeterminate Form $\infty - \infty$

Indeterminate Forms $0^{0}$ , $1^{\pm \infty}$ , $(\pm \infty)^{0}$

Frequently Asked Questions

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L'Hôpital's Rule for Indeterminate Limits

L'Hôpital's Rule for 0 / 0

Examples: Form 0 / 0

Limitations of L'Hôpital's Rule